Question

what is the value of sin 15​

Answer 1

Answer:

$\sin \left(15^{\circ \:}\right)=\dfrac{\sqrt{2-\sqrt{3}}}{2}\quad \begin{pmatrix}\mathrm{Decimal:}&0.25882\dots \end{pmatrix}$

Step-by-step explanation:

$\sin \left(15^{\circ \:}\right)$

$\mathrm{Write}\:\sin \left(15^{\circ \:}\right)\:\mathrm{as}\:\sin \left(\dfrac{30^{\circ \:}}{2}\right)$

$=\sin \left(\dfrac{30^{\circ \:}}{2}\right)$

$\mathrm{Using\:the\:half\:angle\:identity}:\quad \sin \left(\dfrac{x}{2}\right)=\sqrt{\dfrac{1-\cos \left(x\right)}{2}}$

$=\sqrt{\dfrac{1-\cos \left(30^{\circ \:}\right)}{2}}$

$\mathrm{Use\:the\:following\:trivial\:identity}:\quad \cos \left(30^{\circ \:}\right)=\dfrac{\sqrt{3}}{2}$

$=\sqrt{\dfrac{1-\dfrac{\sqrt{3}}{2}}{2}}$

$\sqrt{\dfrac{1-\dfrac{\sqrt{3}}{2}}{2}}$

$\dfrac{1-\dfrac{\sqrt{3}}{2}}{2}$

$\mathrm{Join}\:1-\dfrac{\sqrt{3}}{2}:\quad \dfrac{2-\sqrt{3}}{2}$

$=\dfrac{\dfrac{2-\sqrt{3}}{2}}{2}$

$\mathrm{Apply\:the\:fraction\:rule}:\quad \dfrac{\dfrac{b}{c}}{a}=\dfrac{b}{c\:\cdot \:a}$

$=\dfrac{2-\sqrt{3}}{2\cdot \:2}$

$\mathrm{Multiply\:the\:numbers:}\:2\cdot \:2=4$

$=\dfrac{2-\sqrt{3}}{4}$

$=\sqrt{\dfrac{2-\sqrt{3}}{4}}$

$\mathrm{Apply\:radical\:rule\:}\sqrt[n]{\dfrac{a}{b}}=\dfrac{\sqrt[n]{a}}{\sqrt[n]{b}},\:\quad \mathrm{\:assuming\:}a\ge 0,\:b\ge 0$

$=\dfrac{\sqrt{2-\sqrt{3}}}{\sqrt{4}}$

$\sqrt{4}=2$

$=\dfrac{\sqrt{2-\sqrt{3}}}{2}$

$=\dfrac{\sqrt{2-\sqrt{3}}}{2}$